srm 551

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250

Description

题意：一个长度最多$50$的字符串，每次操作可以交换相邻的两个字符，问，经过最多$MaxSwaps$次交换之后，最多能让多少个相同的字符连起来

Solution

枚举一个字符，让其他字符往这里即可，取最小即可

Code

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#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define F first
#define S second
typedef long long LL;
typedef pair<int, int> pii;
struct ColorfulChocolates {
    int maximumSpread(string chocolates, int maxSwaps) {
        int ans = 0;
        int n = chocolates.size();
        for (int i = 0; i < n; ++i) {
            string s = chocolates;
            int now = 1;
            vector<int> a;
            for (int j = i - 1; j >= 0; --j)
                if (s[j] == s[i])   a.pb(i - j - now), ++now;
            now = 1;
            for (int j = i + 1; j < n; ++j)
                if (s[j] == s[i])   a.pb(j - i - now), ++now;
            sort(a.begin(), a.end());
            int cnt = maxSwaps;
            int t = 1;
            for (int i = 0; i < a.size(); ++i)
                if (cnt - a[i] >= 0) {
                    cnt -= a[i];
                    ++t;
                }
            ans = max(ans, t);
        }
        return ans;
    }
};

450

Description

一个有向图，顶点标号$0\sim n-1$，每个点会选择优先走他能到达的编号最小的点，现在想通过去掉一些边使得可以从$0$走到$n-1$,求最少要去掉的边数

Solution

考虑类似于$floyd$的做法，$d[i][j]$表示$i$走到$j$需要最少删多少条边，直接$dp$即可

Code

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#include <bits/stdc++.h>//dp
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define F first
#define S second
typedef long long LL;
typedef pair<int, int> pii;
const int N = 55, inf = 1e6;
int d[N][N];
struct ColorfulWolves {
    int getmin(vector <string> colormap) {
        int n = colormap.size();
        for (int i = 0; i < n; ++i) {
            int now = 0;
            for (int j = 0; j < n; ++j) {
                if (colormap[i][j] == 'Y')  d[i][j] = now++;
                else d[i][j] = inf;
            }
        }
        for (int k = 0; k < n; ++k)
            for (int i = 0; i < n; ++i)
                for (int j = 0; j < n; ++j)
                    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
        return d[0][n - 1] == inf ? -1 : d[0][n - 1];
    }
};